Split JSON string column to multiple columns

Question

I'm looking for a generic solution to extract all the json fields as columns from a JSON string column.

df =  spark.read.load(path)
df.show()

File format of the files in 'path' is parquet

Sample data

|id | json_data
| 1 | {"name":"abc", "depts":["dep01", "dep02"]}
| 2 | {"name":"xyz", "depts":["dep03"],"sal":100}
| 3 | {"name":"pqr", "depts":["dep02"], "address":{"city":"SF","state":"CA"}}

Expected output

|id | name    | depts              | sal | address_city | address_state
| 1 | "abc"   | ["dep01", "dep02"] | null| null         | null
| 2 | "xyz"   | ["dep03"]          | 100 | null         | null
| 3 | "pqr"   | ["dep02"]          | null| "SF"         | "CA"

I know that I can extract the columns by creating a StructType with the schema defined and using 'from_json' method.

But this approach requires manual schema definition.

val myStruct = StructType(
  Seq(
    StructField("name", StringType),
    StructField("depts", ArrayType(StringType)),
    StructField("sal", IntegerType)
  ))

var newDf = df.withColumn("depts", from_json(col("depts"), myStruct))

Is there a better way to flatten the JSON column without manually defining the schema? In the example provided, I can see the JSON fields available. But in reality, I can't traverse all the rows to find all the fields.

So I'm looking for a solution to split all the fields to columns without specifying the names or types of the columns.

if the data is a pure json multi rows then it is possible to implement the schema automatically. — Daeho Ro
– Daeho Ro, Commented Sep 3, 2019 at 23:24
It is a pure json column but not all the fields are available in all the rows. Like first row in my example is missing 'sal' field — Munesh
– Munesh, Commented Sep 3, 2019 at 23:44

Gaurang Shah · Accepted Answer · 2019-09-05 01:56:42Z

3

If it's a CSV file and only one column is coming as JSON data. You can use following solution.

val csvDF = spark.read.option("delimiter", "|").option("inferSchema", true).option("header", true).csv("test.csv")
val rdd = csvDF.select(" json_data").rdd.map(_.getString(0))
val ds = rdd.toDS
val jsonDF = spark.read.json(ds)
val jsonDFWithID = jsonDF.withColumn("id", monotonically_increasing_id())
val csvDFWithID = csvDF.select($"id ").withColumn("id", monotonically_increasing_id())
val joinDF = jsonDFWithID.join(csvDFWithID, "id").drop("id")

This is how final Data Frame look like.

scala> joinDF.printSchema()
root
 |-- address: struct (nullable = true)
 |    |-- city: string (nullable = true)
 |    |-- state: string (nullable = true)
 |-- depts: array (nullable = true)
 |    |-- element: string (containsNull = true)
 |-- name: string (nullable = true)
 |-- sal: long (nullable = true)
 |-- id : double (nullable = true)

Following solution would work if it's a JSON file. for me. inferSchema works perfectly fine.

json File

~/Downloads ▶ cat test.json
{"id": 1, "name":"abc", "depts":["dep01", "dep02"]},
{"id": 2, "name":"xyz", "depts" :["dep03"],"sal":100}

code

scala> scc.read.format("json").option("inerSchema", true).load("Downloads/test.json").show()
+--------------+---+----+----+
|         depts| id|name| sal|
+--------------+---+----+----+
|[dep01, dep02]|  1| abc|null|
|       [dep03]|  2| xyz| 100|
+--------------+---+----+----+

edited Sep 5, 2019 at 1:56

answered Sep 4, 2019 at 1:55

Gaurang Shah

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3 Comments

Munesh Over a year ago

My data/file format is not JSON. Only one of the column is a JSON string.

Gaurang Shah Over a year ago

@Munesh I have modified the answer. is this what you want?

Munesh Over a year ago

Thanks Gaurang Shah. Your approach helped me with my solution. This approach didn't work with nested JSON like my example and monotonically_increasing_id didn't work because it is not sequential and hence join returns only first few rows.

akuiper · Accepted Answer · 2019-09-04 01:36:27Z

1

Assuming json_data is of type map (which you can always convert to map if it's not), you can use getItem:

df = spark.createDataFrame([
    [1, {"name": "abc", "depts": ["dep01", "dep02"]}],
    [2, {"name": "xyz", "depts": ["dep03"], "sal": 100}]
],
    ['id', 'json_data']
)

df.select(
    df.id, 
    df.json_data.getItem('name').alias('name'), 
    df.json_data.getItem('depts').alias('depts'), 
    df.json_data.getItem('sal').alias('sal')
).show()

+---+----+--------------+----+
| id|name|         depts| sal|
+---+----+--------------+----+
|  1| abc|[dep01, dep02]|null|
|  2| xyz|       [dep03]| 100|
+---+----+--------------+----+

A more dynamic way to extract columns:

cols = ['name', 'depts', 'sal']
df.select(df.id, *(df.json_data.getItem(col).alias(col) for col in cols)).show()

edited Sep 4, 2019 at 1:36

answered Sep 4, 2019 at 1:28

akuiper

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1 Comment

Munesh Over a year ago

I can't specify the column names in the variable 'cols' as I don't know all the fields available in the JSON

Munesh · Accepted Answer · 2019-09-06 18:53:07Z

Based on @Gaurang Shah's answer, I have implemented a solution to handle nested JSON structure and fixed the issues with using monotonically_increasing_id(Non-sequential)

In this approach, 'populateColumnName' function recursively checks for StructType column and populate the column name.

'renameColumns' function renames the columns by replacing '.' with '_' to identify the nested json fields.

'addIndex' function adds index to the dataframe to join the dataframe after parsing the JSON column.

def flattenJSON(df : DataFrame, columnName: String) : DataFrame = {

    val indexCol = "internal_temp_id"

    def populateColumnName(col : StructField) : Array[String] = {
        col.dataType match {
          case struct: StructType => struct.fields.flatMap(populateColumnName).map(col.name + "." + _)
          case rest         => Array(col.name)
        }
    }

    def renameColumns(name : String) : String = {
        if(name contains ".") {
            name + " as " + name.replaceAll("\\.", "_")
        }
        else name
    }

    def addIndex(df : DataFrame) : DataFrame = {

        // Append "rowid" column of type Long
        val newSchema = StructType(df.schema.fields ++ Array(StructField(indexCol, LongType, false)))

        // Zip on RDD level
        val rddWithId = df.rdd.zipWithIndex
        // Convert back to DataFrame
        spark.createDataFrame(rddWithId.map{ case (row, index) => Row.fromSeq(row.toSeq ++ Array(index))}, newSchema)
    }

    val dfWithID = addIndex(df)

    val jsonDF = df.select(columnName)

    val ds = jsonDF.rdd.map(_.getString(0)).toDS
    val parseDF = spark.read.option("inferSchema",true).json(ds)

    val columnNames = parseDF.schema.fields.flatMap(populateColumnName).map(renameColumns)

    var resultDF = parseDF.selectExpr(columnNames:_*)

    val jsonDFWithID = addIndex(resultDF)

    val joinDF = dfWithID.join(jsonDFWithID, indexCol).drop(indexCol)

    joinDF
}

val res = flattenJSON(jsonDF, "address")

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